At `490^(@)C,` the equilibrium constant for the stnthesis of HI is 50, the value of K for the dissocisation of HI will be

Options

A. `20.0`
B. `2.0`
C. `0.2`
D. 0.02`

Explanation

To solve the problem, we need to find the equilibrium constant for the dissociation of HI given the equilibrium constant for the synthesis of HI.

### Step-by-Step Solution:

1. **Identify the Reaction for Synthesis of HI:**
The synthesis of hydrogen iodide (HI) can be represented by the following balanced equation:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]
The equilibrium constant (K) for this reaction at \(490^\circ C\) is given as 50.

2. **Write the Expression for the Equilibrium Constant:**
The equilibrium constant expression for the synthesis reaction is:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Given that \(K_c = 50\).

3. **Reverse the Reaction for Dissociation of HI:**
The dissociation of HI can be represented by the reverse reaction:
\[ 2 HI(g) \rightleftharpoons H_2(g) + I_2(g) \]

4. **Determine the Equilibrium Constant for the Reversed Reaction:**
When a reaction is reversed, the equilibrium constant for the reverse reaction (let's call it \(K'\)) is the reciprocal of the equilibrium constant for the forward reaction:
\[ K' = \frac{1}{K_c} \]
Therefore, substituting the value of \(K_c\):
\[ K' = \frac{1}{50} \]

5. **Calculate the Value of \(K'\):**
Performing the calculation:
\[ K' = 0.02 \]

6. **Conclusion:**
The value of the equilibrium constant for the dissociation of HI at \(490^\circ C\) is:
\[ K' = 0.02 \]

### Final Answer:
The value of \(K\) for the dissociation of HI is **0.02**.